Area of triangle

Let $m\angle A=\theta$ . Then the area of $\Delta ABC=\frac{1}{2}\cdot AB \cdot AC \sin \theta \text{ (equation (1))}$ and the area of $\Delta APQ=\frac{1}{2} \cdot AQ \cdot AP \sin \theta \text{ (equation (2))}$ Using equation (1), we can solve for $\sin \theta$ . Since $AB=14, AC=15$ and $[ABC]=84,$ we have $84=\frac{1}{2}(14)(15)\sin \theta$ So $\sin \theta=\frac{84 \cdot 2}{14 \cdot 15}=\frac{4}{5}.$ Now using equation (2), we can obtain the area of $\Delta APQ$ . With $\sin \theta=\frac{4}{5}, AP=9$ and $AQ=5$ , we have $[APQ]=\frac{1}{2}(5)(9)\left(\frac{4}{5}\right)=\boxed{18}.$

Notice that : $\dfrac{[APQ]}{[ABC]} = \dfrac{\overline{AP}\cdot\overline{AQ}}{\overline{AB}\cdot\overline{AC}}$ . So: $84 \dfrac{ 9\cdot 5}{14\cdot 15} = \boxed{018}$ is the area of APQ.

Exercise: Prove the first relation.

Area of triangle APQ = (1/3) area of triangle APC = (1/3)(9/14)area of triangle ABC = (3/14)(84) = 18

Let the altitude from C to base AB have length h and intersect side AB at point D. $[ABC]=\frac{ 1 }{ 2 }(AB)(h)=84$ $h=12$

Since the altitude is perpendicular to AB by definition, we can find the length BD because leg CD and hypotenuse BC form a standard Pythagorean (5, 12, 13) triple. So D is 5 away from B and thus 9 away from A. This means that D and P are the same point. Now consider the right triangle ACP, which has two legs, AP = 9 and CP = 12, and hypotenuse AC = 15. This means the area of triangle ACP is 54.

No matter where Q is on side AC, triangles APQ and CPQ share the same altitude. Thus the ratio of the two areas of the triangles is proportional to the ratio of the lengths of the bases of the triangles. Since the base AP of triangle APQ is half the base CP of triangle CPQ (because AP is 5 and AC is 15), this means that the area of triangle APQ is one-half the area of triangle CPQ. These two areas must add to the area of triangle ACP, so the area of triangle APQ is one-third of the area of triangle ACP, or 18 .

((15 * 14)/5 ) *2 =84

((9 * 5 )/5)*2 = 18

area=1/2 a b*sin@

C'mon!!! I shouldn't have got the points... I did it manually, using my ruler and compass... : (

Area of triangle ABC = .5 * 15 * h = 84 So, h = 11.2. Since AB = 14 and AP = 9, while P lies on AB, the height 'h' would be reduced proportionally and the new height would be 7.2 Area of triangle APQ = .5 * 5 * 7.2 = 18

Why is area of ABC given if its three sides are given ?

area of triangle ABC=0.5 × AB × hieght=84
hieght=84÷7=12
therefote hieght=cp
perpendicular line from P toAC=9× 12÷15=7.2 (euclids)

area of triangle APQ=0.5 × 5×7.2=18

We can find the angle that both triangles shares by law of cosines, and will give approx. $53.13$ . Then, go to the small triangle. We have two sides and the angle formed by them. Again, law of cosines looking for the other side (call it $c$ ):

$c^{2} = 9^{2} + 5^{2} - 2(9)(5) \cos 53.12$

$c = 2\sqrt {13}$

Apply Heron's Formula:

$[APQ] = \sqrt{(7 + \sqrt{13})(7 + \sqrt{13} - 9)(7 + \sqrt{13} - 5)(7 + \sqrt{13} - 2\sqrt{13})}$

Solving this will give our answer:

$[APQ] = \boxed{18}$ .

Is it an original way?

Area ABC is proportional to product of AB and AC. Area APQ is proportional to product of AP and AQ. Common to these is cosine of angle A. Thus area APQ = 84 9 5/(14*15) = 18 sq. Units.